Hydrolysis calculations: salts of weak acids are bases

Problem #1: What is the pH of a 0.100 M solution of sodium acetate? Kb = 5.65 x 10¯10. (I will use NaAc as shorthand)

Solution:

1) Here is the chemical reaction (net ionic) for the hydrolysis of NaAc:

Ac¯ + H2O <==> HAc + OH¯

2) Here is the Kb expression for Ac¯:

[HAc] [OH¯]

Kb =

----------------

[Ac¯]

3) We can then substitute values into the Kb expression in the normal manner:

(x) (x)

5.65 x 10¯10 =

----------------

0.100 - x

4) Ignoring the minus x in the usual manner, we proceed to solve for the hydroxide ion concentration:

x = square root of [(5.65 x 10¯10) (0.100)]

x = 7.52 x 10¯6 M = [OH¯]

We then calculate the pH. Since this is a base calculation, we need to do the pOH first:

pOH = - log 7.52 x 10¯6 = 5.124

pH = 14 - 5.124 = 8.876

Problem #2: What is the pH of a 0.0500 M solution of KCN? Kb = 2.1 x 10¯5.

Solution:

1) Here is the chemical reaction (net ionic) for the hydrolysis of KCN:

CN¯ + H2O <==> HCN + OH¯

Here is the Kb expression for CN¯:

[HCN] [OH¯]

Kb =

----------------

[CN¯]

2) We can then substitute values into the Kb expression in the normal manner:

(x) (x)

2.1 x 10¯5 =

----------------

0.050 - x

3) Ignoring the minus x in the usual manner, we proceed to sove for the hydroxide ion concentration:

x = square root of [(2.1 x 10¯5) (0.050)]

x = 1.025 x 10¯3 M = [OH¯]

4) We then calculate the pH. Since this is a base calculation, we need to do the pOH first:

pOH = - log 1.025 x 10¯3 = 2.99

pH = 14 - 2.99 = 11.01

I certainly hope that you see that the two calculations are exactly alike, except for the numbers. In fact, when I wrote this tutorial in August 2002, I just copied and pasted the first example and changed the words and numbers. One final point to make: textbooks will often say "find the pH of the cyanide ion, CN¯." What they really mean is a complete chemical substance of which only the CN¯ is chemically relevant to the problem. Please keep in mind that the solution contains a cation (K+ in the example above, Na+ in the first example.